Question

Consider the current carrying loop shown in the figure formed by the combination of radial wires and segment of circle whose center is at point $\text{O}$. Find the magnitude of $\overrightarrow{B}$ at point $\text{O}$.

• A. $\frac{1}{4}[\frac{{\mu }_{0}I}{2a}-\frac{{\mu }_{0}I}{2b}]$
• B. $\frac{1}{2}[\frac{{\mu }_{0}I}{2a}-\frac{{\mu }_{0}I}{2b}]$
• C. $\frac{1}{6}[\frac{{\mu }_{0}I}{2a}-\frac{{\mu }_{0}I}{2b}]$
• D. None of these

Answer 1

The correct option is C $\frac{1}{6}[\frac{{\mu }_{0}I}{2a}-\frac{{\mu }_{0}I}{2b}]$

Magnetic field due to straight wire at $\text{O}$ is zero because point $\text{O}$ is lying on the axis of wire or just near the axis. So magnetic field at point $\text{O}$ will be produced due to circular arc only.

Thus, the magnetic field due to an arc of radius $a$ is given by,

$B_1=\frac{{\mu }_{0}I}{2a}\times \frac{{60}^{\circ }}{{360}^{\circ }}=\frac{{\mu }_{0}I}{2a}\frac{1}{6}$

Now using right- hand thumb rule for the direction of magnetic field

$\therefore B_1=\frac{{\mu }_{0}I}{2a}\frac{1}{6}⨀$

$⨀$ indicates the direction of magnetic field perpendicularly outward to the plane of paper.

Magnetic field due to arc of radius $b$ is,

$B_2=\frac{{\mu }_{0}I}{2b}\times \frac{{60}^{\circ }}{{360}^{\circ }}$

$\therefore B_2=\frac{{\mu }_{0}I}{2b}\times \frac{1}{6}⨂$

So, the net magnetic field at point $\text{O}$ will be

$\left|\begin{array}{c}\overrightarrow{B_{net}}\end{array}\right|=\left|\begin{array}{c}\overrightarrow{B_1}\end{array}\right|-\left|\begin{array}{c}\overrightarrow{B_2}\end{array}\right|$

(considering outward direction to the plane of paper to be positive)

$B_{net}=[\frac{{\mu }_{o}I}{2a}\frac{1}{6}-\frac{{\mu }_{o}I}{2b}\frac{1}{6}]$

$\therefore B_{net}=\frac{1}{6}[\frac{{\mu }_{o}I}{2a}-\frac{{\mu }_{o}I}{2b}]$

Hence, option $\left(c\right)$ is the right answer.